Beiträge zur Algebra und Geometrie
Contributions to Algebra and Geometry
Volume 47 (2006), No. 2, 519-526.
Rank of Matrices and the
Pexider Equation
Józef Kalinowski
Uniwersytet Śla̧ski, Instytut Matematyki
ul. Bankowa 14, 40-007 Katowice, Poland
e-mail: kalinows@ux2.math.us.edu.pl
Abstract. The paper [1] gives a characterization of those linear operators which preserve the rank of matrices over R. In this paper we characterize the operators of type (1) having this property, without making the
linearity assumption. For matrices from Mn,m and min{n, m} ≥ 3 the
operator must be linear and of the form from [1]. If 1 ≤ min{n, m} ≤ 2,
then the operator may be nonlinear.
MSC2000: 15A03
Keywords: rank of matrices
Introduction
Let R denote the set of real numbers and N the set of positive integers. Let
m, n ∈ N be constants. Let Mm,n be the set of m×n real matrices, i.e. M ∈ Rm×n
and Mn = Mn,n , where m, n ∈ N.
First of all let us introduce
Definition 1. We say that an operator
F = [fi ],
where fi : R −→ R for i = 1, 2, . . . , m,
(1)
preserves the rank of matrices from Mm,n if for every matrix A ∈ Mm,n the equation
rank(A) = rank(F (A))
(2)
holds, where the matrix F (A) := [fi (ai,j )] for i = 1, 2, . . . , m; j = 1, 2, . . . , n.
c 2006 Heldermann Verlag
0138-4821/93 $ 2.50 520
J. Kalinowski: Rank of Matrices and the Pexider Equation
The problem of the form of operators defined on the space of matrices over R into
itself was studied by many authors under assumption that the operator is linear
(see references of [1]).
In this paper, using the Pexider type additive functional equation, we obtain
without a linearity assumption some results for the operator of the form (1). We
prove, that if min{m, n} ≥ 3, then a rank preserving operator must be linear and
of the form from paper [1]. If the cases 1 ≤ min{m, n} ≤ 2 the operator may be
nonlinear.
Main results
Remark 1. An operator F of the form (1) preserves the rank of matrices from
M1 if and only if
f1 (x) = 0 ⇐⇒ x = 0.
We prove the following
Lemma 1. Let an operator F of the form (1) be an operator preserving the rank
of matrices from Mn for n ≥ 2. Then the equivalence
x=0
⇐⇒
fi (x) = 0
for
i = 1, 2, . . . , n
(3)
is true.
Proof. For the matrix B1 ∈ Mn with all entries equal to zero, rank(B1 ) = 0. If the
operator F is an operator preserving the rank of matrices, then rank(F (B1 )) = 0
and it follows that fi (x) = 0 for i = 1, 2, . . . , n, i.e.
x=0
=⇒
fi (x) = 0 for i = 1, 2, . . . , n.
(4)
Consider the matrix
B2 = diag (x, x, . . . , x),
where x ∈ R and x 6= 0. Using implication (4) we obtain
F (B2 ) = diag (f1 (x), f2 (x), . . . , fn (x))
for arbitrary x ∈ R.
Let us observe that rank(B2 ) = n. If the operator F is an operator preserving
the rank of matrices from Mn , then rank(F (B2 )) = n and det(F (B2 )) 6= 0, i.e.
f1 (x) · f2 (x) · · · fn (x) 6= 0. Then fi (x) 6= 0 for i = 1, 2, . . . , n.
We obtain the implication
x 6= 0
fi (x) 6= 0 for i = 1, 2, . . . , n.
=⇒
Then the equivalent implication
fi (x) = 0
=⇒
x = 0 for i = 1, 2, . . . , n
(5)
J. Kalinowski: Rank of Matrices and the Pexider Equation
holds.
From (4) and (5) the equivalence (3) is true.
521
We prove
Theorem 1. An operator F of the form (1) is an operator preserving the rank of
matrices from M2 if and only if there exist constants ci 6= 0, i = 1, 2, such that
fi (x) = ci · g(x),
x ∈ R.
(6)
Then the function g is an injective solution of the multiplicative functional Cauchy
equation
g(x · y) = g(x) · g(y) for x, y ∈ R,
(7)
such that
⇐⇒
g(x) = 0
x = 0.
(8)
Proof. Let F be an operator preserving the rank of matrices from M2 and ci =
fi (1) for i = 1, 2. From Lemma 1 we obtain ci 6= 0 for i = 1, 2.
First consider the matrix
x 1
B3 =
x 1
for arbitrary x ∈ R. Then rank(F (B3 )) = 1 and det(F (B3 )) = 0. Now
f1 (x) · c2 = c1 · f2 (x) for x ∈ R.
Define new functions
gi (x) =
fi (x)
ci
(9)
for x ∈ R
for i = 1, 2.
Then from (9) we define the function
g(x) := g1 (x) = g2 (x) for x ∈ R.
Next consider the matrix
B4 =
1 x
y x·y
(10)
for arbitrary x, y ∈ R. Then rank(F (B4 )) = 1 and det(F (B4 )) = 0. Now
c1 · f2 (x · y) = f1 (x) · f2 (y).
Dividing both sides by c1 · c2 6= 0 and using the definition (10) we obtain
g2 (x · y) = g1 (x) · g2 (y) for x, y ∈ R.
(11)
From (10) and (11) the function g satisfies the multiplicative functional Cauchy
equation (7).
522
J. Kalinowski: Rank of Matrices and the Pexider Equation
Consider the matrix
B5 =
x 1
y 1
for arbitrary x, y ∈ R, x 6= y. Then det(B5 ) 6= 0 and rank(B5 ) = 2. Then
rank(F (B5 )) = 2 and det(F (B5 )) 6= 0, i.e. f1 (x) · c2 − f2 (y) · c1 6= 0, then from
(10)
g(x) 6= g(y) for x 6= y, x, y ∈ R.
This means that the function g is injective. Condition (8) follows from Lemma 1.
Let us assume that an operator F is of the form (1), with functions fi for
i = 1, 2. The injective function g fulfils the multiplicative functional Cauchy
equation (7) and the coefficients ci 6= 0 for i = 1, 2. We prove that F is an
operator preserving the rank of matrices from M2 .
Let us consider an arbitrary matrix H ∈ M2 of the form
u v
H=
,
w z
where u, v, w, z ∈ R.
Consider three possible cases:
◦
1 rank(H) = 2:
Then det(H) 6= 0 and u · z − v · w 6= 0, i.e. u · z 6= v · w. From injectivity of the
function g we obtain that g(u·z) 6= g(v·w). The function g fulfils the multiplicative
functional Cauchy equation (7), so we obtain g(u) · g(z) 6= g(v) · g(w). Multiplying
both sides of the above relation by c1 · c2 6= 0 we obtain
c1 · g(u) · c2 · g(z) 6= c1 · g(v) · c2 · g(w)
and from definition (3)
f1 (u) · f2 (z) − f1 (v) · f2 (w) 6= 0.
Then det(F (H)) 6= 0 and rank(F (H)) = 2.
2◦ rank(H) = 1:
Then det(H) = 0 and u · z − v · w = 0, i.e. u · z = v · w. From the injectivity of the
function g we obtain that g(u·z) = g(v·w). The function g fulfils the multiplicative
functional Cauchy equation (4), so we obtain g(u) · g(z) = g(v) · g(w). Multiplying
both sides of the above relation by c1 · c2 6= 0 we obtain
c1 · g(u) · c2 · g(z) = c1 · g(v) · c2 · g(w)
and from definition (3)
f1 (u) · f2 (z) − f1 (v) · f2 (w) = 0.
Thus det(F (H)) = 0. The rank of the matrix H is one, so at least one of the entries
u, v, w, z is nonzero. Then from injectivity of the function g at least one of the
J. Kalinowski: Rank of Matrices and the Pexider Equation
523
numbers g(u), g(v), g(w), g(z) is nonzero. Multiplying these numbers by nonzero
coefficients c1 , c1 , c2 , c2 , respectively, we obtain that at least one of the entries
f1 (u), f1 (v), f2 (w), f2 (z) of the matrix F (H) is nonzero and has rank(F (H)) = 1.
3◦ rank(H) = 0:
Then u = v = w = z = 0. From Lemma 1 and definition (6) we obtain that
f1 (u) = f1 (v) = f2 (w) = f2 (z) = 0 and rank(F (H)) = 0.
A few simple numerical examples will illustrate the role of the injectivity of the
function g in Theorem 1.
Example 1. The operator F defined by formulae f1 (x) = x3 , f2 (x) = 2x3 for
x ∈ R fulfils the assumptions of Theorem 1. It is the nonlinear operator of the
form (1) preserving the rank of matrices from M2 .
Without the assumption of injectivity on the function g Theorem 1 is not true.
Example 2. Let g(x) = x2 , and x ∈ R a non-injective solution of the multiplicative functional Cauchy equation (7). Let f1 (x) = f2 (x) = x2 and consider the
matrices
1 1
1 1
B6 =
and F (B6 ) =
.
−1 1
1 1
Observe that det(B6 ) = 2 and rank(B6 ) = 2, det(F (B6 )) = 0 and rank(F (B6 )) = 1.
The operator F is not an operator preserving the rank of matrices from M2 .
Let us observe that the result obtained in Theorem 1 for n = 2 is not true for
n = 3. Consider the following
Example 3. Let g(x) = x3 , and x ∈ R an injective solution of the functional
multiplicative Cauchy equation (7). Let f1 (x) = f3 (x) = x3 , f2 (x) = 2x3 and
consider the matrices




1 1 0
1 1 0
B7 =  1 0 1  and F (B7 ) =  2 0 2  .
2 1 1
8 1 1
Observe that det(B7 ) = 0 and rank(B7 ) = 2, det(F (B7 )) = 12 and rank(F (B7 )) = 3.
The operator F is not an operator preserving the rank of matrices from M3 .
For n = 3 we prove
Theorem 2. An operator F of the form (1) is an operator preserving the rank of
matrices from M3 if and only if there exist constants ci 6= 0, i = 1, 2, 3 such that
fi (x) = ci · x,
x ∈ R.
(12)
524
J. Kalinowski: Rank of Matrices and the Pexider Equation
Proof. Let F be an operator preserving the rank of matrices from M3 . From
Lemma 1 it follows that the equivalence (3) holds. Let the constants ci := fi (1) 6= 0
for i = 1, 2, 3.
First consider the matrices




x
1 0
f1 (x)
c1 0
0 1  and F (B8 ) =  f2 (y)
0 c2 
B8 =  y
x+y 1 1
f3 (x + y) c3 c3
for arbitrary x, y ∈ R. Then
det(F (B8 )) = c1 · c2 · f3 (x + y) − f1 (x) · c2 · c3 − f2 (y) · c1 · c3 .
Because rank(B8 ) = 2, we have rank(F (B8 )) = 2 and det(F (B8 )) = 0, so the
following equation is satisfied:
c1 · c2 · f3 (x + y) = f1 (x) · c2 · c3 + f2 (y) · c1 · c3
for all x, y ∈ R. From the above and c1 · c2 · c3 6= 0 we obtain
f1 (x) f2 (y)
f3 (x + y)
=
+
c3
c1
c2
for x, y ∈ R. We define new functions
hi (x) :=
1
· fi (x) for x ∈ R,
ci
(13)
where i = 1, 2, 3. In other words, the Pexider type additive functional equation
h3 (x + y) = h1 (x) + h2 (y) for x, y ∈ R
is fulfilled by hi , i = 1, 2, 3.
By [2, Theorem 1, p. 317] the functions
h1 (x) = h(x) + α1 ,
h2 (x) = h(x) + α2 ,
h3 (x) = h(x) + α1 + α2
are the solution, where h is an additive function with constants α1 , α2 ∈ R. For
an additive function h it follows h(0) = 0 and from (3) we obtain α1 = α2 = 0.
Then h1 = h2 = h3 = h and the additive functional Cauchy equation
h(x + y) = h(x) + h(y) for all x, y ∈ R
is fulfilled.
For any square matrix


1 x 0
B9 =  y x · y 0 
0 0 1


c1
f1 (x)
0
and F (B9 ) =  f2 (y) f2 (x · y) 0  ,
0
0
c3
(14)
J. Kalinowski: Rank of Matrices and the Pexider Equation
525
where x, y ∈ R, we obtain rank(B9 ) = 2 and also rank(F (B9 )) = 2. Thus
det(F (B9 )) = c3 ·[c1 ·f2 (x·y)−f1 (x)·f2 (y)] = 0. From the above and c1 ·c2 ·c3 6= 0,
together with h1 = h2 = h, the function h satisfies the multiplicative functional
Cauchy equation
h(x · y) = h(x) · h(y) for all x, y ∈ R.
(15)
Now, from [2; Theorem 1, p. 356], it follows that the only functions satisfying
simultaneously (14) and (15), i.e. the additive and the multiplicative functional
Cauchy equations are h = 0 and h = id, where id denotes the identity function
on R. Since h(1) = 1, we see that in our case h(x) = x for x ∈ R.
From definition (11) we obtain for i = 1, 2, 3 that
fi (x) = ci · hi (x) = ci · h(x) = ci · x for all x ∈ R,
where ci = fi (1) 6= 0 and the functions fi , i = 1, 2, 3, are of the form (13).
From the properties of determinants it follows that the operators F the form
(1) preserve the rank of matrices from M3 .
We prove a theorem which describes all operators of the form (1) preserving the
rank of real matrices from Mn for n ≥ 3.
Theorem 3. An operator F of the form (1) preserves the rank of matrices from
Mn for n ≥ 3 if and only if fi (x) = ci · x, i = 1, 2, . . . , n, where ci 6= 0 are
constants.
Proof. Assume that the operator F preserves the rank of any matrix H from Mn ,
where n ≥ 3.
For n = 3 the assertion was proved in Theorem 2.
For n > 3 consider the matrices Di ∈ Mn with a minor of degree 3 of the form


ai,i
ai,i+1
ai,i+2
Hi =  ai+1,i ai+1,i+1 ai+1,i+2 
ai+2,i ai+2,i+1 ai+2,i+2
for i = 1, 2, . . . , n − 2 and other entries equal to zero. Observe that rank(Di ) =
rank(Hi ). For an operator F preserving the rank of matrices from Mn it follows
that rank(F (Di )) = rank(Fi (Hi )), where Fi = [fi , fi+1 , fi+2 ] for i = 1, 2, . . . , n − 2.
From Theorem 2 we obtain that there exist constants ck 6= 0 such that fk (x) = ck ·x
for k = i, i + 1, i + 2, where k = 1, 2, . . . , n − 2. Then the operator F is of the
form (1).
From the properties of determinants it follows that the operators of the form
(1) preserve the rank of matrices from Mn for n ≥ 3.
Remark 2. If we define an operator F (A) := [fj (ai,j )] for i = 1, 2, . . . , m, j =
1, 2, . . . , n for matrices A ∈ Mm,n , then results analogous to those in Theorems 1,
2 and 3 are valid.
526
J. Kalinowski: Rank of Matrices and the Pexider Equation
Corollary. If m > n, then results analogous to those in Theorems 1, 2 and 3 are
valid for matrices from Mm,n .
Remark 3. The results in Theorems 1, 2 and 3 were obtained without assumptions on the operator F (i.e. continuity, measurability or others).
In the paper [1] a similar problem and result are presented. Let F : Mm,n −→
Mm,n be a linear operator. By [1, Theorem 3.1] we obtain that rank(F (A)) =
rank(A) for all A ∈ Mm,n if and only if there exist invertible matrices M ∈ Mm
and N ∈ Mn such that F (A) = M AN or if m = n then F (A) = M At N , where
At denotes transposition of the matrix A.
Let us observe that using Remark 2 in case min{m, n} ≥ 3 the result obtained
for operators of the form (1) is the same as that in Theorem 3. The M is the
diagonal matrix M = diag (c1 , c2 , . . . , cm ) and N = In , where In ∈ Mn is the unit
matrix.
In case n = 2 we obtain a better result: operators F of the form (1) preserving
the rank of matrices from M2 may be linear or nonlinear (see Example 1).
Acknowledgement. I would like to express my thanks to professors Roman Ger
and Mieczyslaw Kula for their valuable suggestions and remarks.
References
[1] Li, Chi-Kwong; Pierce, S.: Linear Preserver Problems. Am. Math. Monthly
Zbl
0991.15001
108 (2001), 591–605.
−−−−
−−−−−−−−
[2] Kuczma, M.: An introduction to the theory of functional equations and inequalities. Cauchy’s equation and Jensen’s inequality. Uniwersytet Śla̧ski, Katowice,
Polish Sci. Publ., Warsaw 1985.
Zbl
0555.39004
−−−−
−−−−−−−−
Received June 10, 2005